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Force Tension Question

  1. Apr 1, 2007 #1
    [​IMG]
    - The angle between the ramp and horizontal is 60 degrees.
    - The hanging mass has a mass of 1.5 kg
    - The mass on the ramp has a mass of 0.4857 kg
    - When the system is moving, it has an acceleration of 4.915 m/s^2.

    Determine the tension on the string when:
    a) The system is not moving
    b) The system is released and free to move

    a) I believe that the tension force on the rope would be 14.7 N (1.5 kg * 9.8 m/s^2), as the hanging mass is pulling the rope whereas the stationary ramp mass acts as an anchor.
    b)
    [tex]\Sigma F = ma [/tex]
    [tex]\Sigma F = F_(gravity) - F_(tension) [/tex]
    [tex] F_g - F_t = ma [/tex]
    [tex] 14.7 N - F_t = (1.5 kg)(4.915 m/s^2) [/tex]
    [tex] F_t = 7.33 N [/tex]

    I am not too sure about the second answer, as I'm not really certain whether the mass in the 'ma' expression should be the entire system mass, or whether I have actually done the entire thing correctly in the first place...

    Thanks for any help!!
     
    Last edited: Apr 1, 2007
  2. jcsd
  3. Apr 1, 2007 #2

    Dick

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    A somewhat better answer to a) is to repeat the calculation in b) with a=0. You are right to concentrate on the hanging mass. So m is the hanging mass. Nothing else.
     
  4. Apr 1, 2007 #3
    Thank you for your reply.

    I have one further question regarding this:

    If you wish to have the system accelerate with a value of exactly 3.00 m/s^2, what hanging mass would you need to select to have this happen?

    I'm not sure that this would involve just the mass of the hanging weight, as that is the system acceleration (though both masses are obviously acceleration at the same speed). This is my best attempt, though I believe I am quite mistaken:

    [tex]\Sigma F = ma[/tex]
    [tex]F_g - F_t = ma[/tex]
    [tex]9.8m - F_t = (3.00m/s^2)m[/tex]
    [tex]F_t = 6.8m[/tex]

    Though I can't think of any other equation with which I could solve this like a system or how to find tension...

    Thanks for any help!
     
  5. Apr 1, 2007 #4

    Dick

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    You are not mistaken. But you are correct that you can't solve this for m. To do that you need to start solving equations on the ramp side. Hint, there are three forces acting on the ramp mass, T, gravity and friction. You should be able to determine a coefficient of kinetic friction from the first problem and then go from there.
     
  6. Apr 1, 2007 #5
    Thanks! That makes sense, as the kinetic friction was involved in an earlier question. Thus:

    [tex]\Sigma F = F_t - F_k - F_gh[/tex]
    [tex]ma = F_t - F_k - F_gh[/tex]
    [tex](3.00 m/s^2)m = F_t - 0.809 N - 4.122281 N [/tex]
    [tex]3m = F_t - 4.931281[/tex]
    [tex]F_t = 6.8m [/tex]
    [tex]3m = 6.8m - 4.931281[/tex]
    [tex]m = 1.644 kg [/tex]

    Those would be the calculations for an acceleration of 3.00 m/s^2 up the ramp, whereas I believe the calculations for the acceleration down the ramp would include the addition of the kinetic friction force instead of its subtraction.

    Thanks again!
     
  7. Apr 1, 2007 #6

    Dick

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    You're welcome. But recheck your calculation. You have the counterintuitive result that mass went up but the acceleration went down. That can't be right.
     
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